If one real root of the quadratic equation is cube of the other root, then a value of is
- A.100
- B.144
- C.
- D.
Show answer and solution
Correct option: D
Solution
Let the roots be α and α³. By Vieta's formulas:
Product: α · α³ = α⁴ = 256/81, so α = ±4/3.
Sum: α + α³ = -k/81.
Taking α = -4/3: α³ = -64/27.
Sum = -4/3 + (-64/27) = -36/27 - 64/27 = -100/27.
So -k/81 = -100/27 → k = 81 × 100/27 = 300.
Taking α = 4/3: α³ = 64/27.
Sum = 4/3 + 64/27 = 36/27 + 64/27 = 100/27.
So -k/81 = 100/27 → k = -300.
Thus k = -300.
Product: α · α³ = α⁴ = 256/81, so α = ±4/3.
Sum: α + α³ = -k/81.
Taking α = -4/3: α³ = -64/27.
Sum = -4/3 + (-64/27) = -36/27 - 64/27 = -100/27.
So -k/81 = -100/27 → k = 81 × 100/27 = 300.
Taking α = 4/3: α³ = 64/27.
Sum = 4/3 + 64/27 = 36/27 + 64/27 = 100/27.
So -k/81 = 100/27 → k = -300.
Thus k = -300.
[JEE Main 2019 · 11 Jan · Shift I]